The Correct Answer is Option (C)

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy.

Kinetic energy = $\frac{1}{2}m{v}^2$

Potential energy = $\frac{KQq}{r}$

$\therefore$ $\frac{1}{2}m{v}^2$ = $\frac{KQq}{r}$

$\Rightarrow$ r $\propto$ $\frac{1}{m}$

The Correct Answer is Option (C)

Kinetic energy of alpha nucleus is equall to electrostatic potential energy of the system of the alpha particle and the heavy nucleus. That is,

$\frac{1}{2}m{v}^2=\frac{1}{4\pi {\epsilon }_0}\times \frac{\left(Ze\right)\times \left(2e\right)}{d_{in}}$

$\therefore$ d_min $\propto$ $\frac{1}{m}$